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3 years ago

When light waves hit ice, most of them bounce off and radiate back into space. Because of this, it can be said that ice is a(n)

Physics

2 answers:

Softa [21]3 years ago

8 0

Reflective~~~~~~~~~~~~~~~~~~~~~~

kirill [66]3 years ago

8 0

Answer;

Reflective surface

Explanation;

Reflection is one of the property of a wave that occurs when a wave hits a boundary which is a reflective surface. The wave bounces off upon hitting the reflective boundary, such that the angle of incidence is equal to the angle of reflection.
Therefore, ice is an example of a reflective surface as light waves bounces off and radiate back when they hit the surface of the ice.

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A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

What determines an object’s thermal energy

Tems11 [23]

Answer:

temperature and mass

Explanation:

The higher the temperature of a given quantity of a substance, more is its thermal energy.

If a substance contains more mass, this also implies that the object has more particles in it . hence, it has high thermal energy.

Additional Information :

Temperature is a measure of the average kinetic energy of the particles of a substance.

The thermal energy of an object depends on three factors:

number of molecules in the object
temperature of the object.
thermal energy it has.

3 0

2 years ago

A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an

nekit [7.7K]

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

3 0

2 years ago

Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat

Nataly_w [17]

Answer:

$\omega_{f} = 1000000\,\frac{rev}{day}$

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

$\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}$

$\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}$

$\omega_{f} = 1000000\,\frac{rev}{day}$

3 0

3 years ago

A ferry is travelling at 10 m/s relative to the river towards the south. The river is flowing at 2 m/s relative to

mixer [17]

Answer:

11 m/s south

Explanation:

The velocity of the passenger relative to the river bank is equal to the velocity of the passenger relative to the ferry, plus the velocity of the ferry relative to the river, plus the velocity of the river relative to the river bank.

v_passenger,bank = v_passenger,ferry + v_ferry,river + v_river,bank

If we take north to be positive and south to be negative:

v = 1.0 m/s + (-10 m/s) + (-2 m/s)

v = -11 m/s

v = 11 m/s south

8 0

3 years ago